Hello!

Please, any good soul to help me with:

$\displaystyle \intx^2sin2xdx$???

The result is :

$\displaystyle \frac{1}{4}(1-2x^2)cos2x+\frac{1}{2}xsin2x+C$

Many thanks!!!

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- Jan 17th 2013, 08:43 AM #1

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- Jan 17th 2013, 09:42 AM #2

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## Re: \int

You left out the space between 'int' and 'x^2'. I added it.

Use "integration by parts": $\displaystyle \int u dv= uv- \int v du$ with $\displaystyle u= x^2$, $\displaystyle dv= sin(2x)dx$ so that $\displaystyle du= 2xdx$, $\displaystyle v= -\frac{1}{2}cos(2x)$. Doing that, you will still have an integral involving $\displaystyle xcos(2x)$. So use integration by parts**again**, this time letting $\displaystyle u= x$, $\displaystyle dv= cos(2x)dx$.

Generally speaking, any time you have an integral of $\displaystyle x^n f(x)dx$, you can consider repeated integration by parts, repeatedly taking the power of x as "u" so that when you differentiate to get du, you have one less power of x. Repeat until you have reduced the power of x to 0.

The result is :

$\displaystyle \frac{1}{4}(1-2x^2)cos2x+\frac{1}{2}xsin2x+C$

Many thanks!!!